Establish the absolute maximum of a function

Question

We have this function:$$f(x)=\begin{cases} \sin(x) \cdot\ln(\sin2x), & \mbox{if }0<x<\pi/2 \\ 0, & \mbox{if }x=0,\mbox{or }x=\pi/2 \end{cases}$$

So, how to prove that it decreases and increases in $]0,\pi/4[$? I think we can't study the sign of the derivative in an analytic way. Plotting it you can see that it admits even the absolute maximum; in fact, for each $0<x<\pi/2$ we have:$$\sin(x) \cdot\ln(\sin2x)\le0$$ how you can verify quickly. So here we notice that the absolute maximum is $0$. Suppose you don't see that $f(x)\le0$ for each $x$ in that interval; so can you study that function in some way? Can you find the absolute maximum using the derivative?

Answer 1

Let's find the derivative: we have $$ g(x) = \cos(x)\ln(\sin 2x) + \sin x \frac{2 \cos 2x}{\sin 2x}\\ = \cos(x) \ln(\sin 2x) + 2\sin x \cot(2x) $$ In order to find where the function has relative extrema, set this equal to zero $$ \cos(x) \ln(\sin 2x) + 2\sin x \cot(2x) = 0 \implies\\ \cos(x) \ln(\sin 2x) = -2\sin x \cot(2x) \implies\\ -\ln(\sin 2x) = 2\tan(x)\cot(2x) \implies\\ -\ln(\sin 2x) = 1 - \tan^2 x $$ Admittedly, that last manipulation is a bit tricky, but the point is that we only need to show that the two sides go from having the same sign to having opposite signs.

Answer 2

Because $f$ is differentiable in the open interval $(0,\tfrac \pi 2)$ you can use the derivative as normal to study the monotonicity of $f$ in this open interval.

To study the monotonicity of $f$ in the whole domain $[0,\tfrac \pi 2]$ you have to examine the values of $f(0)$ and $f\left(\tfrac \pi2\right)$ in comparision to the values of the rest of the domain. You will see (in the following $x_0$ is the smallest zero of the derivative in the domain $(0,\tfrac \pi 2)$ which is already calculated by Omnomnomnom):

• Because $f(0)$ is bigger than $f(x)$ for any $x\in (0,x_0]$ and because $f$ is decreasing in $(0,\tfrac \pi4]$ (derivative has negative sign), $f$ is decreasing in $[0,x_0]$.
• $f$ is increasing in $[x_0,\tfrac \pi4]$ because the sign of the derivative is positive there.
• $f$ is decreasing in $[\tfrac \pi4,\tfrac \pi2)$ but not in $[\tfrac \pi4,\tfrac \pi2]$ because $f\left(\tfrac \pi2\right)=0$ is bigger than any $f(x)$ for $x\in (\tfrac \pi4,\tfrac \pi2)$